Question #115447
A basketball is shot at 12.0 m/s at a 55.0 degree angle. What is the velocity of the ball 2.16 seconds later?
1
Expert's answer
2020-05-14T09:12:51-0400

The ball has two projections of speed: vxv_{x} and vyv_{y} .

At the initial moment:

v0x=v0×cos(α)v0y=v0×sin(α)v_{0x}=v_{0}\times \cos(\alpha) \newline v_{0y}=v_{0}\times \sin(\alpha)

The x-projection of speed will not change in time.

The y-projection:

vy=v0y+at=v0ygtv_{y}=v_{0y}+at=v_{0y}-gt

Take g=10g=10 :

vx=v0x=12×cos(55)6.88vy=12×sin(55)10×2.1611.76v_{x}=v_{0x}=12\times\cos(55)\approx6.88\newline v_{y}=12\times\sin(55)-10\times2.16\approx-11.76

So:

v=vx2+vy2=47.3344+138.297613.63m/sv=\sqrt{v_{x}^2+v_{y}^2}=\sqrt{47.3344+138.2976}\approx13.63m/s


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